Software Tools for Earth and Environmental Science
– 9th WEEK –
Emir Toker
14/11/2019
R Promgramming - Part 1
- Syllabus and Book
- DataCamp Class
- Repeat - R Language
- Practice - R Language
- Practice - Create a Function
Coffee Break
- R Programming - Conditions
- Comparison Operators
- if Statement
- else Statement
- ifelse Statement
- Next Week
(Optional) Practice - MidTerm Project
Syllabus and Book
Syllabus
Extended Syllabus PDF
Book
PDF - (179 - 185)
DataCamp - Class
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R Language - Repeat
R Language - Repeat
- Create/Open R Project or R File
- Basic Math, Assigment, Comment
- Data Types - Classes
- Numeric
- Integer
- Logical
- Character
- Data Structures - Objects
- Vector
- Matrice
- Array
- Data Frame
- List
- Special Values, Attributes
Create/Open R Project or R File
Project
File - New Project - New Directory
Project
New Project
Project
Directory Name - Create Project
NEW Project is Ready
File
File - New File - Script
Basic Math, Assigment, Comment
Basic Math, Assigment, Comment
Arithmetic Operators ( + - * / ^ )
Scientific Math
Assignment Arrow (<-)
a <- 3.8- Comment Lines, Comment Out, Hash (#)
# value <- old_code()
value <- new_code()Data Types - Classes
Data Types - Classes
- Numeric
- Integer
- Logical
- Character
- Complex
- Date
a <- 4
b <- 3:9
c <- 7L
d <- 1i
e <- 5 < -9
f <- "23"
date <- "2007-06-22"
date1 <- as.Date(date)
date2 <- as.Date("2004-02-13")
date1 - date2
## Time difference of 1225 daysCoercion
Data Structures - Objects
Data Structures - Objects
Data Structures - Vector
c()
spring_months <- c("March", "April", "May", "June")
spring_months[2] <- "new"
spring_months## [1] "March" "new" "May" "June"myvec1 <- c(1, 3, 1, 42)
a <- 35
myvec2 <- c(3L, myvec1, 1e+03, 64^0.5, 2+(3-1.1)/9.44, a)
myvec2## [1] 3.000000 1.000000 3.000000 1.000000 42.000000 1000.000000
## [7] 8.000000 2.201271 35.000000?lenght ?seq ?round ?rep ?sort ?runif ?set.seed
Data Structures - Matrice
matrix(data = ,nrow = ,ncol = )
data <- runif(9,1,100)
data## [1] 12.915285 93.166266 91.116730 30.696196 11.418564 98.591149 3.229152
## [8] 95.188369 52.501424matrix <- matrix(data, nrow = 3, ncol = 3)
matrix## [,1] [,2] [,3]
## [1,] 12.91528 30.69620 3.229152
## [2,] 93.16627 11.41856 95.188369
## [3,] 91.11673 98.59115 52.501424print(c(length(matrix),dim(matrix)))## [1] 9 3 3Data Structures - Array
array(data = ,dim = )
data <- 1:24
array <- array(data, dim = c(4,3,2)) # raw, col, level
array## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24array[2,2,1] # raw, col, level## [1] 6Data Structures - Factor
gender = c("male", "female", "male", "male", "female")
gender## [1] "male" "female" "male" "male" "female"class(gender)## [1] "character"str(gender)## chr [1:5] "male" "female" "male" "male" "female"gender[2]## [1] "female"Data Structures - Factor
gender_factor <- factor(c("male", "female", "male", "male", "female"))
gender_factor## [1] male female male male female
## Levels: female maleclass(gender_factor)## [1] "factor"str(gender_factor)## Factor w/ 2 levels "female","male": 2 1 2 2 1gender_factor[2]## [1] female
## Levels: female maleData Structures - Data Frame
data.frame(data1,data2,data3…)
person=c("Peter", "Lois", "Meg", "Chris", "Stewie")
age=c(42, 40, 17, 14 ,1)
sex=factor(c("M", "F", "F", "M", "M"))
married=c(TRUE, TRUE, FALSE, FALSE, FALSE)
df <- data.frame(person, age, sex, married)
df## person age sex married
## 1 Peter 42 M TRUE
## 2 Lois 40 F TRUE
## 3 Meg 17 F FALSE
## 4 Chris 14 M FALSE
## 5 Stewie 1 M FALSEstr(df)## 'data.frame': 5 obs. of 4 variables:
## $ person : Factor w/ 5 levels "Chris","Lois",..: 4 2 3 1 5
## $ age : num 42 40 17 14 1
## $ sex : Factor w/ 2 levels "F","M": 2 1 1 2 2
## $ married: logi TRUE TRUE FALSE FALSE FALSEData Structures - Data Frame
data.frame(data1,data2,data3…)
person=c("Peter", "Lois", "Meg", "Chris", "Stewie")
age=c(42, 40, 17, 14 ,1)
sex=factor(c("M", "F", "F", "M", "M"))
married=c(TRUE, TRUE, FALSE, FALSE, FALSE)
df <- data.frame(person ,age, sex, married, stringsAsFactors=FALSE)
df## person age sex married
## 1 Peter 42 M TRUE
## 2 Lois 40 F TRUE
## 3 Meg 17 F FALSE
## 4 Chris 14 M FALSE
## 5 Stewie 1 M FALSEstr(df)## 'data.frame': 5 obs. of 4 variables:
## $ person : chr "Peter" "Lois" "Meg" "Chris" ...
## $ age : num 42 40 17 14 1
## $ sex : Factor w/ 2 levels "F","M": 2 1 1 2 2
## $ married: logi TRUE TRUE FALSE FALSE FALSEData Structures - Data Frame
data.frame(data1,data2,data3…)
df[1]## person
## 1 Peter
## 2 Lois
## 3 Meg
## 4 Chris
## 5 Stewiedf[[1]] # df$person## [1] "Peter" "Lois" "Meg" "Chris" "Stewie"df[[1]][1]## [1] "Peter"Data Structures - List
list(data1,data2,data3…)
matrix <- matrix(data=1:4,nrow=2,ncol=2)
vector <- c(T,F,T,T)
var <- "hello"
data_frame <- data.frame(person ,age, sex, married, stringsAsFactors=FALSE)
list <- list(matrix,vector,var,data_frame)
list## [[1]]
## [,1] [,2]
## [1,] 1 3
## [2,] 2 4
##
## [[2]]
## [1] TRUE FALSE TRUE TRUE
##
## [[3]]
## [1] "hello"
##
## [[4]]
## person age sex married
## 1 Peter 42 M TRUE
## 2 Lois 40 F TRUE
## 3 Meg 17 F FALSE
## 4 Chris 14 M FALSE
## 5 Stewie 1 M FALSESpecial Values, Attributes
Special Values
NA, NaN, NULL, Inf
class(NA) # Not Available (“missing” entity)## [1] "logical"class(NaN) # Not a Number## [1] "numeric"class(NULL) # Null (“empty” entity)## [1] "NULL"class(Inf) # Infinity## [1] "numeric"Attributes
person=c("Peter", "Lois", "Meg", "Chris", "Stewie")
age=c(42, 40, 17, 14 ,1)
sex=factor(c("M", "F", "F", "M", "M"))
married=c(TRUE, TRUE, FALSE, FALSE, FALSE)
data_frame <- data.frame(person ,age, sex, married, stringsAsFactors=FALSE)
data_frame## person age sex married
## 1 Peter 42 M TRUE
## 2 Lois 40 F TRUE
## 3 Meg 17 F FALSE
## 4 Chris 14 M FALSE
## 5 Stewie 1 M FALSEdata_frame[2]## age
## 1 42
## 2 40
## 3 17
## 4 14
## 5 1Attributes
attributes(data_frame)## $names
## [1] "person" "age" "sex" "married"
##
## $class
## [1] "data.frame"
##
## $row.names
## [1] 1 2 3 4 5attr(data_frame,"row.names") <- c("bir", "iki", "uc", "dort","bes")
data_frame## person age sex married
## bir Peter 42 M TRUE
## iki Lois 40 F TRUE
## uc Meg 17 F FALSE
## dort Chris 14 M FALSE
## bes Stewie 1 M FALSEPractice - R Language
Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
mydata <- read.csv(file = "18397_Cekmekoy_Omerli_15dk.txt",
header = TRUE,
sep = ";")
mydata## sta_no year month day hour minutes temp precipitation pressure
## 1 18397 2017 7 26 18 0 23.9 0.00 1003.0
## 2 18397 2017 7 26 18 15 23.9 0.00 1003.1
## 3 18397 2017 7 26 18 30 23.8 0.00 1003.2
## 4 18397 2017 7 26 18 45 23.8 0.00 1003.2
## 5 18397 2017 7 26 19 0 23.6 0.00 1003.2
## 6 18397 2017 7 26 19 15 23.2 0.00 1003.1
## 7 18397 2017 7 26 19 30 23.2 0.00 1003.1
## 8 18397 2017 7 26 19 45 23.1 0.00 1003.1
## 9 18397 2017 7 26 20 0 23.0 0.00 1003.1
## 10 18397 2017 7 26 20 15 22.8 0.00 1003.0
## 11 18397 2017 7 26 20 30 22.5 0.00 1003.0
## 12 18397 2017 7 26 20 45 22.4 0.00 1003.0
## 13 18397 2017 7 26 21 0 22.2 0.00 1003.0
## 14 18397 2017 7 26 21 15 22.3 0.00 1003.0
## 15 18397 2017 7 26 21 30 22.2 0.00 1003.1
## 16 18397 2017 7 26 21 45 21.7 0.00 1003.1
## 17 18397 2017 7 26 22 0 21.9 0.00 1003.2
## 18 18397 2017 7 26 22 15 21.7 0.00 1003.3
## 19 18397 2017 7 26 22 30 21.6 0.00 1003.3
## 20 18397 2017 7 26 22 45 22.2 0.00 1003.4
## 21 18397 2017 7 26 23 0 22.2 0.00 1003.4
## 22 18397 2017 7 26 23 15 22.1 0.00 1003.5
## 23 18397 2017 7 26 23 30 22.3 0.00 1003.4
## 24 18397 2017 7 26 23 45 22.5 0.00 1003.4
## 25 18397 2017 7 27 0 0 22.3 0.00 1003.4
## 26 18397 2017 7 27 0 15 22.2 0.00 1003.2
## 27 18397 2017 7 27 0 30 22.5 0.00 1003.2
## 28 18397 2017 7 27 0 45 22.6 0.00 1003.2
## 29 18397 2017 7 27 1 0 22.6 0.00 1003.3
## 30 18397 2017 7 27 1 15 22.6 0.00 1003.4
## 31 18397 2017 7 27 1 30 22.6 0.00 1003.2
## 32 18397 2017 7 27 1 45 22.7 0.00 1003.2
## 33 18397 2017 7 27 2 0 22.6 0.00 1003.3
## 34 18397 2017 7 27 2 15 22.5 0.00 1003.2
## 35 18397 2017 7 27 2 30 22.6 0.00 1003.2
## 36 18397 2017 7 27 2 45 22.5 0.00 1003.1
## 37 18397 2017 7 27 3 0 22.5 0.00 1003.1
## 38 18397 2017 7 27 3 15 22.4 0.00 1003.0
## 39 18397 2017 7 27 3 30 22.5 0.00 1003.1
## 40 18397 2017 7 27 3 45 22.4 0.00 1003.3
## 41 18397 2017 7 27 4 0 22.5 0.00 1003.4
## 42 18397 2017 7 27 4 15 22.6 0.00 1003.5
## 43 18397 2017 7 27 4 30 23.0 0.00 1003.5
## 44 18397 2017 7 27 4 45 23.2 0.00 1003.5
## 45 18397 2017 7 27 5 0 24.2 0.00 1003.6
## 46 18397 2017 7 27 5 15 25.1 0.00 1003.5
## 47 18397 2017 7 27 5 30 25.5 0.00 1003.4
## 48 18397 2017 7 27 5 45 26.1 0.00 1003.3
## 49 18397 2017 7 27 6 0 27.1 0.00 1003.3
## 50 18397 2017 7 27 6 15 26.9 0.00 1003.3
## 51 18397 2017 7 27 6 30 27.6 0.00 1003.3
## 52 18397 2017 7 27 6 45 28.0 0.00 1003.2
## 53 18397 2017 7 27 7 0 28.4 0.00 1003.1
## 54 18397 2017 7 27 7 15 28.5 0.00 1003.1
## 55 18397 2017 7 27 7 30 29.3 0.00 1003.0
## 56 18397 2017 7 27 7 45 30.2 0.00 1002.9
## 57 18397 2017 7 27 8 0 30.1 0.00 1002.8
## 58 18397 2017 7 27 8 15 30.1 0.00 1002.8
## 59 18397 2017 7 27 8 30 30.4 0.00 1002.8
## 60 18397 2017 7 27 8 45 30.4 0.00 1002.8
## 61 18397 2017 7 27 9 0 30.8 0.00 1002.9
## 62 18397 2017 7 27 9 15 30.9 0.00 1002.8
## 63 18397 2017 7 27 9 30 31.0 0.00 1002.6
## 64 18397 2017 7 27 9 45 31.5 0.00 1002.6
## 65 18397 2017 7 27 10 0 31.2 0.00 1002.6
## 66 18397 2017 7 27 10 15 30.9 0.00 1002.4
## 67 18397 2017 7 27 10 30 30.9 0.00 1002.4
## 68 18397 2017 7 27 10 45 30.4 0.00 1002.3
## 69 18397 2017 7 27 11 0 30.4 0.00 1002.1
## 70 18397 2017 7 27 11 15 30.0 0.00 1001.9
## 71 18397 2017 7 27 11 30 29.2 0.00 1001.9
## 72 18397 2017 7 27 11 45 29.5 0.00 1001.7
## 73 18397 2017 7 27 12 0 29.4 0.00 1001.6
## 74 18397 2017 7 27 12 15 29.3 0.00 1001.3
## 75 18397 2017 7 27 12 30 29.6 0.00 1001.2
## 76 18397 2017 7 27 12 45 28.8 0.00 1001.3
## 77 18397 2017 7 27 13 0 29.0 0.00 1001.1
## 78 18397 2017 7 27 13 15 29.0 0.00 1001.2
## 79 18397 2017 7 27 13 30 29.2 0.00 1001.3
## 80 18397 2017 7 27 13 45 28.4 0.00 1001.5
## 81 18397 2017 7 27 14 0 27.8 0.00 1001.6
## 82 18397 2017 7 27 14 15 27.4 0.00 1001.6
## 83 18397 2017 7 27 14 30 26.6 0.00 1001.5
## 84 18397 2017 7 27 14 45 26.2 0.00 1001.2
## 85 18397 2017 7 27 15 0 25.8 0.00 1001.1
## 86 18397 2017 7 27 15 15 25.6 0.00 1001.0
## 87 18397 2017 7 27 15 30 25.4 0.00 1000.9
## 88 18397 2017 7 27 15 45 24.2 0.00 1001.8
## 89 18397 2017 7 27 16 0 19.2 7.01 1003.7
## 90 18397 2017 7 27 16 15 19.5 15.81 1003.2
## 91 18397 2017 7 27 16 30 20.1 16.06 1003.1
## 92 18397 2017 7 27 16 45 20.8 16.06 1003.7
## 93 18397 2017 7 27 17 0 21.2 17.19 -9999.0
## 94 18397 2017 7 27 17 15 21.4 17.21 1005.6
## 95 18397 2017 7 27 17 30 21.4 18.46 1005.4
## 96 18397 2017 7 27 17 45 21.4 21.21 1005.1
## 97 18397 2017 7 27 18 0 21.2 21.21 1005.1
## 98 18397 2017 7 27 18 15 21.0 21.21 -9999.0
## 99 18397 2017 7 27 18 30 20.8 21.21 1006.3
## 100 18397 2017 7 27 18 45 20.9 21.21 -9999.0
## 101 18397 2017 7 27 19 0 20.8 21.40 1005.7
## 102 18397 2017 7 27 19 15 20.7 21.40 1006.2
## 103 18397 2017 7 27 19 30 20.8 21.60 1003.6
## 104 18397 2017 7 27 19 45 20.8 21.82 1003.7
## 105 18397 2017 7 27 20 0 20.9 21.82 -9999.0
## 106 18397 2017 7 27 20 15 20.6 21.82 -9999.0
## 107 18397 2017 7 27 20 30 20.6 21.82 1005.1
## 108 18397 2017 7 27 20 45 20.5 21.82 1005.6
## 109 18397 2017 7 27 21 0 20.7 21.82 1005.5
## 110 18397 2017 7 27 21 15 20.8 21.82 1005.7
## 111 18397 2017 7 27 21 30 20.4 21.82 1005.6
## 112 18397 2017 7 27 21 45 20.4 21.82 1005.8
## 113 18397 2017 7 27 22 0 20.6 21.82 1005.8
## 114 18397 2017 7 27 22 15 20.5 21.82 1005.9
## 115 18397 2017 7 27 22 30 20.4 21.82 1006.0
## 116 18397 2017 7 27 22 45 20.5 21.82 1005.9
## 117 18397 2017 7 27 23 0 20.5 21.82 1005.9
## 118 18397 2017 7 27 23 15 20.6 21.82 1005.9
## 119 18397 2017 7 27 23 30 20.5 21.82 1006.0
## 120 18397 2017 7 27 23 45 20.5 21.82 1006.0
## 121 18397 2017 7 28 0 0 20.4 21.82 1006.0
## relative_humidity
## 1 94
## 2 95
## 3 96
## 4 96
## 5 96
## 6 97
## 7 97
## 8 98
## 9 98
## 10 98
## 11 98
## 12 99
## 13 99
## 14 99
## 15 99
## 16 99
## 17 99
## 18 99
## 19 99
## 20 100
## 21 100
## 22 100
## 23 100
## 24 100
## 25 100
## 26 100
## 27 100
## 28 100
## 29 100
## 30 100
## 31 100
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## 33 100
## 34 100
## 35 100
## 36 100
## 37 100
## 38 100
## 39 100
## 40 100
## 41 100
## 42 100
## 43 100
## 44 100
## 45 100
## 46 97
## 47 84
## 48 82
## 49 79
## 50 78
## 51 78
## 52 76
## 53 76
## 54 75
## 55 73
## 56 65
## 57 57
## 58 60
## 59 53
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## 64 53
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## 66 57
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## 70 61
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## 76 70
## 77 68
## 78 69
## 79 69
## 80 71
## 81 72
## 82 72
## 83 77
## 84 79
## 85 80
## 86 82
## 87 84
## 88 79
## 89 99
## 90 100
## 91 100
## 92 100
## 93 100
## 94 100
## 95 100
## 96 100
## 97 100
## 98 100
## 99 100
## 100 100
## 101 100
## 102 100
## 103 100
## 104 100
## 105 100
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## 108 100
## 109 100
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## 111 100
## 112 100
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## 116 100
## 117 100
## 118 100
## 119 100
## 120 100
## 121 100Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Check the class and structure of your new data.
## [1] "data.frame"## 'data.frame': 121 obs. of 10 variables:
## $ sta_no : int 18397 18397 18397 18397 18397 18397 18397 18397 18397 18397 ...
## $ year : int 2017 2017 2017 2017 2017 2017 2017 2017 2017 2017 ...
## $ month : int 7 7 7 7 7 7 7 7 7 7 ...
## $ day : int 26 26 26 26 26 26 26 26 26 26 ...
## $ hour : int 18 18 18 18 19 19 19 19 20 20 ...
## $ minutes : int 0 15 30 45 0 15 30 45 0 15 ...
## $ temp : num 23.9 23.9 23.8 23.8 23.6 23.2 23.2 23.1 23 22.8 ...
## $ precipitation : num 0 0 0 0 0 0 0 0 0 0 ...
## $ pressure : num 1003 1003 1003 1003 1003 ...
## $ relative_humidity: int 94 95 96 96 96 97 97 98 98 98 ...## $names
## [1] "sta_no" "year" "month"
## [4] "day" "hour" "minutes"
## [7] "temp" "precipitation" "pressure"
## [10] "relative_humidity"
##
## $class
## [1] "data.frame"
##
## $row.names
## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
## [18] 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
## [35] 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
## [52] 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68
## [69] 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85
## [86] 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102
## [103] 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119
## [120] 120 121Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Take the “Temperature” parameter and assign it as a new variable.
temp_data <- mydata$temp
temp_data## [1] 23.9 23.9 23.8 23.8 23.6 23.2 23.2 23.1 23.0 22.8 22.5 22.4 22.2 22.3
## [15] 22.2 21.7 21.9 21.7 21.6 22.2 22.2 22.1 22.3 22.5 22.3 22.2 22.5 22.6
## [29] 22.6 22.6 22.6 22.7 22.6 22.5 22.6 22.5 22.5 22.4 22.5 22.4 22.5 22.6
## [43] 23.0 23.2 24.2 25.1 25.5 26.1 27.1 26.9 27.6 28.0 28.4 28.5 29.3 30.2
## [57] 30.1 30.1 30.4 30.4 30.8 30.9 31.0 31.5 31.2 30.9 30.9 30.4 30.4 30.0
## [71] 29.2 29.5 29.4 29.3 29.6 28.8 29.0 29.0 29.2 28.4 27.8 27.4 26.6 26.2
## [85] 25.8 25.6 25.4 24.2 19.2 19.5 20.1 20.8 21.2 21.4 21.4 21.4 21.2 21.0
## [99] 20.8 20.9 20.8 20.7 20.8 20.8 20.9 20.6 20.6 20.5 20.7 20.8 20.4 20.4
## [113] 20.6 20.5 20.4 20.5 20.5 20.6 20.5 20.5 20.4Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
Plot the “temperature” vector.
plot(temp_data)
Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Print minimum temperature and find which element is the minimum in temperature vector.
print(min(temp_data))## [1] 19.2which(temp_data==min(temp_data))## [1] 89which(temp_data==19.2)## [1] 89Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- change the minimum value with NA and Print.
temp_data[which(temp_data==min(temp_data))] <- NA
temp_data[which(temp_data==19.2)] <- NA
temp_data[89] <- NA
print(temp_data)## [1] 23.9 23.9 23.8 23.8 23.6 23.2 23.2 23.1 23.0 22.8 22.5 22.4 22.2 22.3
## [15] 22.2 21.7 21.9 21.7 21.6 22.2 22.2 22.1 22.3 22.5 22.3 22.2 22.5 22.6
## [29] 22.6 22.6 22.6 22.7 22.6 22.5 22.6 22.5 22.5 22.4 22.5 22.4 22.5 22.6
## [43] 23.0 23.2 24.2 25.1 25.5 26.1 27.1 26.9 27.6 28.0 28.4 28.5 29.3 30.2
## [57] 30.1 30.1 30.4 30.4 30.8 30.9 31.0 31.5 31.2 30.9 30.9 30.4 30.4 30.0
## [71] 29.2 29.5 29.4 29.3 29.6 28.8 29.0 29.0 29.2 28.4 27.8 27.4 26.6 26.2
## [85] 25.8 25.6 25.4 24.2 NA 19.5 20.1 20.8 21.2 21.4 21.4 21.4 21.2 21.0
## [99] 20.8 20.9 20.8 20.7 20.8 20.8 20.9 20.6 20.6 20.5 20.7 20.8 20.4 20.4
## [113] 20.6 20.5 20.4 20.5 20.5 20.6 20.5 20.5 20.4Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Plot the new “temperature” vector.
Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Replace these new temperature values with old temperature values located in your data frame.
mydata$temp <- temp_data
mydata## sta_no year month day hour minutes temp precipitation pressure
## 1 18397 2017 7 26 18 0 23.9 0.00 1003.0
## 2 18397 2017 7 26 18 15 23.9 0.00 1003.1
## 3 18397 2017 7 26 18 30 23.8 0.00 1003.2
## 4 18397 2017 7 26 18 45 23.8 0.00 1003.2
## 5 18397 2017 7 26 19 0 23.6 0.00 1003.2
## 6 18397 2017 7 26 19 15 23.2 0.00 1003.1
## 7 18397 2017 7 26 19 30 23.2 0.00 1003.1
## 8 18397 2017 7 26 19 45 23.1 0.00 1003.1
## 9 18397 2017 7 26 20 0 23.0 0.00 1003.1
## 10 18397 2017 7 26 20 15 22.8 0.00 1003.0
## 11 18397 2017 7 26 20 30 22.5 0.00 1003.0
## 12 18397 2017 7 26 20 45 22.4 0.00 1003.0
## 13 18397 2017 7 26 21 0 22.2 0.00 1003.0
## 14 18397 2017 7 26 21 15 22.3 0.00 1003.0
## 15 18397 2017 7 26 21 30 22.2 0.00 1003.1
## 16 18397 2017 7 26 21 45 21.7 0.00 1003.1
## 17 18397 2017 7 26 22 0 21.9 0.00 1003.2
## 18 18397 2017 7 26 22 15 21.7 0.00 1003.3
## 19 18397 2017 7 26 22 30 21.6 0.00 1003.3
## 20 18397 2017 7 26 22 45 22.2 0.00 1003.4
## 21 18397 2017 7 26 23 0 22.2 0.00 1003.4
## 22 18397 2017 7 26 23 15 22.1 0.00 1003.5
## 23 18397 2017 7 26 23 30 22.3 0.00 1003.4
## 24 18397 2017 7 26 23 45 22.5 0.00 1003.4
## 25 18397 2017 7 27 0 0 22.3 0.00 1003.4
## 26 18397 2017 7 27 0 15 22.2 0.00 1003.2
## 27 18397 2017 7 27 0 30 22.5 0.00 1003.2
## 28 18397 2017 7 27 0 45 22.6 0.00 1003.2
## 29 18397 2017 7 27 1 0 22.6 0.00 1003.3
## 30 18397 2017 7 27 1 15 22.6 0.00 1003.4
## 31 18397 2017 7 27 1 30 22.6 0.00 1003.2
## 32 18397 2017 7 27 1 45 22.7 0.00 1003.2
## 33 18397 2017 7 27 2 0 22.6 0.00 1003.3
## 34 18397 2017 7 27 2 15 22.5 0.00 1003.2
## 35 18397 2017 7 27 2 30 22.6 0.00 1003.2
## 36 18397 2017 7 27 2 45 22.5 0.00 1003.1
## 37 18397 2017 7 27 3 0 22.5 0.00 1003.1
## 38 18397 2017 7 27 3 15 22.4 0.00 1003.0
## 39 18397 2017 7 27 3 30 22.5 0.00 1003.1
## 40 18397 2017 7 27 3 45 22.4 0.00 1003.3
## 41 18397 2017 7 27 4 0 22.5 0.00 1003.4
## 42 18397 2017 7 27 4 15 22.6 0.00 1003.5
## 43 18397 2017 7 27 4 30 23.0 0.00 1003.5
## 44 18397 2017 7 27 4 45 23.2 0.00 1003.5
## 45 18397 2017 7 27 5 0 24.2 0.00 1003.6
## 46 18397 2017 7 27 5 15 25.1 0.00 1003.5
## 47 18397 2017 7 27 5 30 25.5 0.00 1003.4
## 48 18397 2017 7 27 5 45 26.1 0.00 1003.3
## 49 18397 2017 7 27 6 0 27.1 0.00 1003.3
## 50 18397 2017 7 27 6 15 26.9 0.00 1003.3
## 51 18397 2017 7 27 6 30 27.6 0.00 1003.3
## 52 18397 2017 7 27 6 45 28.0 0.00 1003.2
## 53 18397 2017 7 27 7 0 28.4 0.00 1003.1
## 54 18397 2017 7 27 7 15 28.5 0.00 1003.1
## 55 18397 2017 7 27 7 30 29.3 0.00 1003.0
## 56 18397 2017 7 27 7 45 30.2 0.00 1002.9
## 57 18397 2017 7 27 8 0 30.1 0.00 1002.8
## 58 18397 2017 7 27 8 15 30.1 0.00 1002.8
## 59 18397 2017 7 27 8 30 30.4 0.00 1002.8
## 60 18397 2017 7 27 8 45 30.4 0.00 1002.8
## 61 18397 2017 7 27 9 0 30.8 0.00 1002.9
## 62 18397 2017 7 27 9 15 30.9 0.00 1002.8
## 63 18397 2017 7 27 9 30 31.0 0.00 1002.6
## 64 18397 2017 7 27 9 45 31.5 0.00 1002.6
## 65 18397 2017 7 27 10 0 31.2 0.00 1002.6
## 66 18397 2017 7 27 10 15 30.9 0.00 1002.4
## 67 18397 2017 7 27 10 30 30.9 0.00 1002.4
## 68 18397 2017 7 27 10 45 30.4 0.00 1002.3
## 69 18397 2017 7 27 11 0 30.4 0.00 1002.1
## 70 18397 2017 7 27 11 15 30.0 0.00 1001.9
## 71 18397 2017 7 27 11 30 29.2 0.00 1001.9
## 72 18397 2017 7 27 11 45 29.5 0.00 1001.7
## 73 18397 2017 7 27 12 0 29.4 0.00 1001.6
## 74 18397 2017 7 27 12 15 29.3 0.00 1001.3
## 75 18397 2017 7 27 12 30 29.6 0.00 1001.2
## 76 18397 2017 7 27 12 45 28.8 0.00 1001.3
## 77 18397 2017 7 27 13 0 29.0 0.00 1001.1
## 78 18397 2017 7 27 13 15 29.0 0.00 1001.2
## 79 18397 2017 7 27 13 30 29.2 0.00 1001.3
## 80 18397 2017 7 27 13 45 28.4 0.00 1001.5
## 81 18397 2017 7 27 14 0 27.8 0.00 1001.6
## 82 18397 2017 7 27 14 15 27.4 0.00 1001.6
## 83 18397 2017 7 27 14 30 26.6 0.00 1001.5
## 84 18397 2017 7 27 14 45 26.2 0.00 1001.2
## 85 18397 2017 7 27 15 0 25.8 0.00 1001.1
## 86 18397 2017 7 27 15 15 25.6 0.00 1001.0
## 87 18397 2017 7 27 15 30 25.4 0.00 1000.9
## 88 18397 2017 7 27 15 45 24.2 0.00 1001.8
## 89 18397 2017 7 27 16 0 NA 7.01 1003.7
## 90 18397 2017 7 27 16 15 19.5 15.81 1003.2
## 91 18397 2017 7 27 16 30 20.1 16.06 1003.1
## 92 18397 2017 7 27 16 45 20.8 16.06 1003.7
## 93 18397 2017 7 27 17 0 21.2 17.19 -9999.0
## 94 18397 2017 7 27 17 15 21.4 17.21 1005.6
## 95 18397 2017 7 27 17 30 21.4 18.46 1005.4
## 96 18397 2017 7 27 17 45 21.4 21.21 1005.1
## 97 18397 2017 7 27 18 0 21.2 21.21 1005.1
## 98 18397 2017 7 27 18 15 21.0 21.21 -9999.0
## 99 18397 2017 7 27 18 30 20.8 21.21 1006.3
## 100 18397 2017 7 27 18 45 20.9 21.21 -9999.0
## 101 18397 2017 7 27 19 0 20.8 21.40 1005.7
## 102 18397 2017 7 27 19 15 20.7 21.40 1006.2
## 103 18397 2017 7 27 19 30 20.8 21.60 1003.6
## 104 18397 2017 7 27 19 45 20.8 21.82 1003.7
## 105 18397 2017 7 27 20 0 20.9 21.82 -9999.0
## 106 18397 2017 7 27 20 15 20.6 21.82 -9999.0
## 107 18397 2017 7 27 20 30 20.6 21.82 1005.1
## 108 18397 2017 7 27 20 45 20.5 21.82 1005.6
## 109 18397 2017 7 27 21 0 20.7 21.82 1005.5
## 110 18397 2017 7 27 21 15 20.8 21.82 1005.7
## 111 18397 2017 7 27 21 30 20.4 21.82 1005.6
## 112 18397 2017 7 27 21 45 20.4 21.82 1005.8
## 113 18397 2017 7 27 22 0 20.6 21.82 1005.8
## 114 18397 2017 7 27 22 15 20.5 21.82 1005.9
## 115 18397 2017 7 27 22 30 20.4 21.82 1006.0
## 116 18397 2017 7 27 22 45 20.5 21.82 1005.9
## 117 18397 2017 7 27 23 0 20.5 21.82 1005.9
## 118 18397 2017 7 27 23 15 20.6 21.82 1005.9
## 119 18397 2017 7 27 23 30 20.5 21.82 1006.0
## 120 18397 2017 7 27 23 45 20.5 21.82 1006.0
## 121 18397 2017 7 28 0 0 20.4 21.82 1006.0
## relative_humidity
## 1 94
## 2 95
## 3 96
## 4 96
## 5 96
## 6 97
## 7 97
## 8 98
## 9 98
## 10 98
## 11 98
## 12 99
## 13 99
## 14 99
## 15 99
## 16 99
## 17 99
## 18 99
## 19 99
## 20 100
## 21 100
## 22 100
## 23 100
## 24 100
## 25 100
## 26 100
## 27 100
## 28 100
## 29 100
## 30 100
## 31 100
## 32 100
## 33 100
## 34 100
## 35 100
## 36 100
## 37 100
## 38 100
## 39 100
## 40 100
## 41 100
## 42 100
## 43 100
## 44 100
## 45 100
## 46 97
## 47 84
## 48 82
## 49 79
## 50 78
## 51 78
## 52 76
## 53 76
## 54 75
## 55 73
## 56 65
## 57 57
## 58 60
## 59 53
## 60 52
## 61 51
## 62 51
## 63 50
## 64 53
## 65 52
## 66 57
## 67 58
## 68 59
## 69 60
## 70 61
## 71 65
## 72 66
## 73 67
## 74 66
## 75 68
## 76 70
## 77 68
## 78 69
## 79 69
## 80 71
## 81 72
## 82 72
## 83 77
## 84 79
## 85 80
## 86 82
## 87 84
## 88 79
## 89 99
## 90 100
## 91 100
## 92 100
## 93 100
## 94 100
## 95 100
## 96 100
## 97 100
## 98 100
## 99 100
## 100 100
## 101 100
## 102 100
## 103 100
## 104 100
## 105 100
## 106 100
## 107 100
## 108 100
## 109 100
## 110 100
## 111 100
## 112 100
## 113 100
## 114 100
## 115 100
## 116 100
## 117 100
## 118 100
## 119 100
## 120 100
## 121 100Practice - R Language
- Read and assign your csv data (Header or seperator ?). “18397_Cekmekoy_Omerli_15dk.txt”
- Check the class and structure of your new data.
- Take the “Temperature” parameter and assign it as a new variable.
- Plot the “temperature” vector.
- Print minimum temperature and find which element is the minimum in temperature vector.
- change the minimum value with NA and Print.
- Plot the new “temperature” vector.
- Replace these new temperature values with old temperature values located in your data frame.
- Write your data frame as a new csv file.
Practice - R Language
- Write your data frame as a new csv file.
write.csv(mydata, file = "new_data.csv")
Practice - Create a Function
Practice - Create a Function
Problem: Take a sample belonged to population, and sum
pop <- 1:6 # This is my population
pop## [1] 1 2 3 4 5 6samp <- sample(pop, size = 2) # This is my sample, I choose two var.
samp## [1] 4 2sum(samp)## [1] 6Practice - Create a Function
I want to create a new function named roll
my_new_function <- function() {
new_variable_1 <- # number or something
new_variable_2 <- # number or something
do_this()
}roll <- function() {
pop <- 1:6
samp <- sample(pop, size = 2)
print(samp)
sum(samp)
}roll()## [1] 4 3## [1] 7Practice - Create a Function
Problem: I want to assign a population spontaneously.
roll_2 <- function() {
pop <-
samp <- sample(pop, size = 2)
print(samp)
sum(samp)
}roll_2() # This will give error. Because pop in undefined.
roll_2 <- function(pop) {
samp <- sample(pop, size = 2)
print(samp)
sum(samp)
}
roll_2(pop = 1:27)## [1] 4 17## [1] 21Practice - Create a Function
You can add new options. { } and () are important
sum(1:27)## [1] 378# Think about these functions
# mean(), print(), plot(), max(), install.packages(), help(), ...R Programming - Conditions
R Programming - Conditions
- Comparison Operators
- equal (==)
- not equal (!=)
- greater or equal to (>=)
- lower or equal to (<=)
- Logical Operators
- the and operator (&)
- the or operator (|)
- the not operator (!)
- if (Stand-Alone) Statement
- else Statement
- ifelse Statement
TRUE & TRUE
TRUE & FALSE
TRUE | FALSE
!TRUE
2 == 3
5 < 6
c(1,4) >= 6
9 != 8
5 < 6 & 9 != 8score <- 80
exam_no <- 2
score >= 75 | exam_no == 1
score>=75 & score<90 | exam_no==1if (Stand-Alone) Statement
The if statement executes a chunk of code if and only if a defined condition is TRUE, which looks something like this:
if(TRUE) message("It was true!")## It was true!if(FALSE) message("It wasn't true!")a <- 3
num <- 4
if ( a <= num ) {
a <- a ^ 2
}
a## [1] 9if (Stand-Alone) Statement
num <- -1
if ( num < 0 ) {
print("num is negative.")
print("Don't worry, I'll fix it.")
num <- num * -1
print("Now num is positive.")
}## [1] "num is negative."
## [1] "Don't worry, I'll fix it."
## [1] "Now num is positive."num## [1] 1if (Stand-Alone) Statement
if (Stand-Alone) Statement
else - Statement
If you want something different to happen when the condition is FALSE, you can add an else declaration.
if(FALSE)
{
message("This won't execute...")
} else
{
message("but this will.")
}## but this will.a <- 3.5
dec <- 0.5
if (dec <= 0.5) {
adec <- dec + 1
} else {
adec <- dec
}
adec## [1] 1.5-else - and -else if- Statement
If your situation has more than two mutually exclusive cases, use else and if statements together.
a <- 1
b <- 1
if (a > b) {
print("A wins!")
} else if (a < b) {
print("B wins!")
} else {
print("Tie.")
}## [1] "Tie."Nesting and Stacking Statements
An if statement can be placed in another if statement. In the editor, modify the mynumber example once more as follows:
Nesting and Stacking Statements
Nesting and Stacking Statements
Nesting and Stacking Statements
Nesting and Stacking Statements
Practice - Conditions - TRUE and FALSE
- Print this if it is TRUE. “This message will print!”
if (...) {
print(...)
} - Go to sleep if it is TRUE, else wake up.
if (...) {
print("Go to sleep!")
} else {
print("Wake up!")
}- The variable message to “I execute this when true!” when the condition is TRUE.
if (...) {
message <- "..."
print(message)
} else {
message <- "I execute this when false!"
print(...)
}Practice - Conditions - Logical Op
Problem : You are a CAR, and you are going on the road, BUT ;
Car_Stop_Light <- 'orange'
Number_of_Pedestrians <- 2Remember rules
if (...) {
print(...);
} else {
print(...);
}Practice - Conditions - Logical Op
ANSWER : You are a CAR, and you are going on the road, BUT ;
Car_Stop_Light <- 'orange'
Number_of_Pedestrians <- 2Remember rules
if (Car_Stop_Light == 'green' & Number_of_Pedestrians ==0) {
print('Go!');
} else {
print('STOP');
}## [1] "STOP"Practice - Conditions - Logical Op
Problem : You want to enjoy, and let’s say the day is;
day <- "Saturday"It is okay, you can fun if it is weekend.
if (...) {
print('Enjoy the weekend!')
} else {
print('Do some work.')
}Practice - Conditions - Logical Op
ANSWER : You want to enjoy, and let’s say the day is;
day <- "Saturday"It is okay, you can fun if it is weekend.
if (day == 'Saturday' | day == 'Sunday') {
print('Enjoy the weekend!')
} else {
print('Do some work.')
}## [1] "Enjoy the weekend!"Practice - Conditions - Logical Op
Problem : You want to go out and your question is “Should I take an umbrella?”
- Check to see, if “weather” is “equal” to “cloudy” and, whether there is a “high_chance_of_rain”.
- If it is both, the code should assign the value of the variable message to be “Take umbrella!”
- Otherwise, the code should assign the value of the variable message to “No need for umbrella!”
- Print the message variable after the conditional statement.
- Based on the condition, what should its value be?
Note : There are two variables in your code, “weather” and “high_chance_of_rain”
Practice - Conditions - Logical Op
# you want to go out and your question is "Should I take an umbrella?"
message <- 'Should I take an umbrella?'
weather <- "cloudy"
high_chance_of_rain <- TRUEPractice - Conditions - Logical Op
# you want to go out and your question is "Should I take an umbrella?"
message <- 'Should I take an umbrella?'
weather <- "cloudy"
high_chance_of_rain <- TRUEif (...) {
message <- 'Take umbrella!'
print(message)
} else {
message <- 'No need for umbrella!'
print(message)
}Practice - Conditions - Logical Op
# you want to go out and your question is "Should I take an umbrella?"
message <- 'Should I take an umbrella?'
weather <- "cloudy"
high_chance_of_rain <- TRUE
if (weather == "cloudy" & high_chance_of_rain == TRUE) {
message <- 'Take umbrella!'
print(message)
} else {
message <- 'No need for umbrella!'
print(message)
}## [1] "Take umbrella!"Next Week
Next Week
Conditions and Loops, DataCamp